Question

If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitric acid, HNO₃(aq) are produced?

Answer 1

17.1195 grams of nitric acid are produced.

Step-by-step explanation:

`3NO_2+H_2O\rightarrow 2HNO_3+NO`

Moles of nitrogen dioxide :

`(25.0 g)/(56 g/mol)=0.5434 mol`

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

`(2)/(3)* 0.5434 mol=0.3623 mol` of nitric acid.

Mass of 0.3623 moles of nitric acid :

`0.3623 mol* 63 g/mol=22.8260 g`

Theoretical yield = 22.8260 g

Experimental yield = ?

`\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}* 100`

`75\%=\frac{\text{Experimental yield}}{22.8260 g}`

Experimental yield of nitric acid = 17.1195 g